What is the angle between #<-4,3,-8 ># and #< 1,-1,5>#?

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Answer 1 · 2016-10-08T14:06:32

#theta ~~ 2.56 radians or 146.76°#

The dot product is used, because it has two definitions:

#barA•barB = (A_1)(B_1) + (A_2)(B_2) + (A_3)(B_3)#

and

#barA•barB = |barA||barB|cos(theta)#

where #theta# is the angle between the two vectors.

Compute the dot product, using the first definition:

#barA•barB = (-4)(1) + (3)(-1) + (-8)(5) = -41#

Compute the magnitude of vector A:

#|barA| = sqrt((-4)² + 3² + (-8)²) = sqrt(89)#

Compute the magnitude of vector B:

#|barB| = sqrt(1² + (-1)² + 5²) = sqrt(27)#

Substitute the above into the second equation:

#-41 = (sqrt(89))(sqrt(27))cos(theta)#

Solve for #theta#:

#theta = cos^-1(-41/(sqrt(89)sqrt(27)))#

#theta ~~ 2.56 radians or 146.76°#